Question

What is the sum of this geometric series?

Answer 1

Answer:

I think D

Step-by-step explanation:

Answer 2

Answer:

D

Step-by-step explanation:

$\sum\limits_{k=1}^38(\frac{1}{4})^{k-1}$

The symbol '$\sum$' , read as sigma, is the symbol for summation. Since the expression below sigma is 'k= 1', while the number above sigma is 3, we are to find the sum of $8(\frac{1}{4})^{k-1}$ with each other from k =1 to k =3.

$\boxed{\text{Multiple rule}: \sum\limits_{i=1}^nku_i=k\sum\limits_{i=1}^nu_i}$

$\sum\limits_{k=1}^38(\frac{1}{4})^{k-1}$

= $8\sum\limits_{k=1}^3(\frac{1}{4})^{k-1}$           (Applying multiple rule)

= $8[(\frac{1}{4})^{1-1}+(\frac{1}{4})^{2-1}+(\frac{1}{4})^{3-1}]$

= $8[(\frac{1}{4})^0+(\frac{1}{4})^1+(\frac{1}{4})^2]$

= $8(1+\frac{1}{4}+\frac{1}{16})$

= $8(\frac{21}{16})$

= $\bf{\frac{21}{2} }$

Thus, the answer is D.